tanC=tan[π-(A+B)]=-tan(A+B)=-(tanA+tanB)/(1-tanAtanB)=-1
则C角最大=135度,B角最小
则由正弦定理:
AB/SinC=AC/sinB => AB/AC=SinC/SinB
SinC= √2/2
SinB=1/√10
最长边与最短边之比= √5
tanC=tan[π-(A+B)]=-tan(A+B)=-(tanA+tanB)/(1-tanAtanB)=-1
则C角最大=135度,B角最小
则由正弦定理:
AB/SinC=AC/sinB => AB/AC=SinC/SinB
SinC= √2/2
SinB=1/√10
最长边与最短边之比= √5